Question

An air-filled parallel-plate capacitor is to be constructed which can store 12 µC of charge when operated at 1200 V. What can be the minimum plate area of the capacitor? The dielectric strength of air is $3\times {10}^6\mathrm{V}{\mathrm{m}}^{-1}.$

Answer 1

Charge that the capacitor can hold = 12 μC
Operating voltage = 1200 V
Breakdown strength, b = $3\times {10}^6\mathrm{V}{\mathrm{m}}^{-1}$
The separation between the plates of the capacitor is given by
$d=\frac{V}{b}=\frac{1200}{3\times {10}^6}=4\times {10}^{-4}\mathrm{m}$
The capacitance of the capacitor is given by
$C=\frac{Q}{V}=\frac{12\times {10}^{-6}}{1200}={10}^{-8}\mathrm{F}$
The capacitance can be expressed in terms of the area of the plates (A) and the separation of the plates (d) as:
$C=\frac{{\in }_{0}A}{d}={10}^{-8}\mathrm{F}$
Thus, the area of the plates is given by
$A=\frac{{10}^{-8}\times (4\times {10}^{-4})}{8.854\times {10}^{-12}}=0.45{\mathrm{m}}^2$